Question

An alpha-particle has a charge of +2e and a mass of 6.64 * 10^-27 kg. It is accelerated from rest through a potential difference that has a value of 1.10 * 10^6 V and then enters a uniform magnetic field whose magnitude is 2.80 T. The alpha-particle moves perpendicular to the magnetic field at all times. What is the speed of the alpha-particle What is the magnitude of the magnetic force on it What is the radius of its circular path

Answer 1

Data Given Charge on the alpha particle $q = 3.2\times 10^{-19} kg$ , Mass of the alpha particle $m= 6.64\times 10^{-27} C$

Potential Difference $V= 1.10\times 10^{6} V$ And Magnetic field

(a) For the speed of alpha particle

We know that $P.D = \frac{Wrok Done}{Charge}\Rightarrow Work Done = P.D\times Charge$

From work energy theorem Work done = Change in K.E

So $K.E. = \frac{1}{2}m.v^{2} = P.D\times q$

$v^{2} = \frac{P.D\times q}{m}= \frac{2\times 1.1\times 10^{6}\times 3.2\times 10^{-19}}{6.64\times 10^{-27}}$ $v^{2} = \frac{P.D\times q}{m}= \frac{2\times 1.1\times 10^{6}\times 3.2\times 10^{-19}}{6.64\times 10^{-27}}= 1.06\times 10^{14}$

$v = \sqrt{1.06\times 10^{14}} = 1.03\times 10^{7} m/s$

(b)

Magnetic Force on the alpha particle is given by

$F_{B} = Bqv\sin \theta = B.q.v= 2.80\times 3.2\times 10^{-19}\times 1.03\times 10^{7}= 9.22\times 10^{-12} N$

(c) Radius of Circular Path is given by

$R = \frac{m.v}{B.q}= \frac{6.64\times 10^{-27}\times 1.03\times 10^{7}}{2.8\times 3.2\times 10^{-19}} =0.7633 \times 10^{1}= 7.633 m$